Integrand size = 87, antiderivative size = 25 \[ \int \frac {1+e^4 \log (x)+\left (6+6 e^4+\left (1+7 e^4\right ) \log (x)+e^4 \log ^2(x)\right ) \log \left (\frac {5+\log (2)}{6+\log (x)}\right )}{\left (6 x+\left (x+6 e^4 x\right ) \log (x)+e^4 x \log ^2(x)\right ) \log \left (\frac {5+\log (2)}{6+\log (x)}\right )} \, dx=\log \left (\frac {x+e^4 x \log (x)}{\log \left (\frac {5+\log (2)}{6+\log (x)}\right )}\right ) \]
Time = 0.43 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.12 \[ \int \frac {1+e^4 \log (x)+\left (6+6 e^4+\left (1+7 e^4\right ) \log (x)+e^4 \log ^2(x)\right ) \log \left (\frac {5+\log (2)}{6+\log (x)}\right )}{\left (6 x+\left (x+6 e^4 x\right ) \log (x)+e^4 x \log ^2(x)\right ) \log \left (\frac {5+\log (2)}{6+\log (x)}\right )} \, dx=6+\log (x)+\log \left (1+e^4 \log (x)\right )-\log \left (\log \left (\frac {5+\log (2)}{6+\log (x)}\right )\right ) \]
Integrate[(1 + E^4*Log[x] + (6 + 6*E^4 + (1 + 7*E^4)*Log[x] + E^4*Log[x]^2 )*Log[(5 + Log[2])/(6 + Log[x])])/((6*x + (x + 6*E^4*x)*Log[x] + E^4*x*Log [x]^2)*Log[(5 + Log[2])/(6 + Log[x])]),x]
Time = 0.92 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {3039, 6, 6, 7239, 7293, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\left (e^4 \log ^2(x)+\left (1+7 e^4\right ) \log (x)+6 e^4+6\right ) \log \left (\frac {5+\log (2)}{\log (x)+6}\right )+e^4 \log (x)+1}{\left (6 x+e^4 x \log ^2(x)+\left (6 e^4 x+x\right ) \log (x)\right ) \log \left (\frac {5+\log (2)}{\log (x)+6}\right )} \, dx\) |
\(\Big \downarrow \) 3039 |
\(\displaystyle \int \frac {e^4 \log \left (\frac {5+\log (2)}{\log (x)+6}\right ) \log ^2(x)+7 e^4 \log \left (\frac {5+\log (2)}{\log (x)+6}\right ) \log (x)+\log \left (\frac {5+\log (2)}{\log (x)+6}\right ) \log (x)+e^4 \log (x)+6 e^4 \log \left (\frac {5+\log (2)}{\log (x)+6}\right )+6 \log \left (\frac {5+\log (2)}{\log (x)+6}\right )+1}{(\log (x)+6) \left (e^4 \log (x)+1\right ) \log \left (\frac {5+\log (2)}{\log (x)+6}\right )}d\log (x)\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {e^4 \log \left (\frac {5+\log (2)}{\log (x)+6}\right ) \log ^2(x)+7 e^4 \log \left (\frac {5+\log (2)}{\log (x)+6}\right ) \log (x)+\log \left (\frac {5+\log (2)}{\log (x)+6}\right ) \log (x)+e^4 \log (x)+\left (6+6 e^4\right ) \log \left (\frac {5+\log (2)}{\log (x)+6}\right )+1}{(\log (x)+6) \left (e^4 \log (x)+1\right ) \log \left (\frac {5+\log (2)}{\log (x)+6}\right )}d\log (x)\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {e^4 \log \left (\frac {5+\log (2)}{\log (x)+6}\right ) \log ^2(x)+\left (1+7 e^4\right ) \log \left (\frac {5+\log (2)}{\log (x)+6}\right ) \log (x)+e^4 \log (x)+\left (6+6 e^4\right ) \log \left (\frac {5+\log (2)}{\log (x)+6}\right )+1}{(\log (x)+6) \left (e^4 \log (x)+1\right ) \log \left (\frac {5+\log (2)}{\log (x)+6}\right )}d\log (x)\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {e^4 \log (x)+(\log (x)+6) \left (e^4 (\log (x)+1)+1\right ) \log \left (\frac {5+\log (2)}{\log (x)+6}\right )+1}{(\log (x)+6) \left (e^4 \log (x)+1\right ) \log \left (\frac {5+\log (2)}{\log (x)+6}\right )}d\log (x)\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {e^4 \log (x)+e^4+1}{e^4 \log (x)+1}+\frac {1}{(\log (x)+6) \log \left (\frac {5+\log (2)}{\log (x)+6}\right )}\right )d\log (x)\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \log (x)+\log \left (e^4 \log (x)+1\right )-\log \left (\log \left (\frac {5+\log (2)}{\log (x)+6}\right )\right )\) |
Int[(1 + E^4*Log[x] + (6 + 6*E^4 + (1 + 7*E^4)*Log[x] + E^4*Log[x]^2)*Log[ (5 + Log[2])/(6 + Log[x])])/((6*x + (x + 6*E^4*x)*Log[x] + E^4*x*Log[x]^2) *Log[(5 + Log[2])/(6 + Log[x])]),x]
3.2.66.3.1 Defintions of rubi rules used
Int[(u_.)*((v_.) + (a_.)*(Fx_) + (b_.)*(Fx_))^(p_.), x_Symbol] :> Int[u*(v + (a + b)*Fx)^p, x] /; FreeQ[{a, b}, x] && !FreeQ[Fx, x]
Int[u_, x_Symbol] :> With[{lst = FunctionOfLog[Cancel[x*u], x]}, Simp[1/lst [[3]] Subst[Int[lst[[1]], x], x, Log[lst[[2]]]], x] /; !FalseQ[lst]] /; NonsumQ[u]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 2.11 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04
method | result | size |
risch | \(\ln \left (x \right )+\ln \left (\ln \left (x \right )+{\mathrm e}^{-4}\right )-\ln \left (\ln \left (\ln \left (x \right )+6\right )-\ln \left (\ln \left (2\right )+5\right )\right )\) | \(26\) |
norman | \(\ln \left (x \right )-\ln \left (\ln \left (\frac {\ln \left (2\right )+5}{\ln \left (x \right )+6}\right )\right )+\ln \left ({\mathrm e}^{4} \ln \left (x \right )+1\right )\) | \(27\) |
parallelrisch | \(-12-\ln \left (\ln \left (\frac {\ln \left (2\right )+5}{\ln \left (x \right )+6}\right )\right )+\ln \left (\left ({\mathrm e}^{4} \ln \left (x \right )+1\right ) {\mathrm e}^{-4}\right )+\ln \left (x \right )\) | \(33\) |
default | \(-\left (-1+\frac {\ln \left (\frac {1}{\ln \left (x \right )+6}\right )}{\ln \left (x \right )+6}\right ) \left (\ln \left (x \right )+6\right )+\ln \left (\frac {6 \,{\mathrm e}^{4}}{\ln \left (x \right )+6}-{\mathrm e}^{4}-\frac {1}{\ln \left (x \right )+6}\right )-\ln \left (\ln \left (\ln \left (2\right )+5\right )+\ln \left (\frac {1}{\ln \left (x \right )+6}\right )\right )\) | \(64\) |
int(((exp(4)*ln(x)^2+(7*exp(4)+1)*ln(x)+6*exp(4)+6)*ln((ln(2)+5)/(ln(x)+6) )+exp(4)*ln(x)+1)/(x*exp(4)*ln(x)^2+(6*x*exp(4)+x)*ln(x)+6*x)/ln((ln(2)+5) /(ln(x)+6)),x,method=_RETURNVERBOSE)
Time = 0.28 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \[ \int \frac {1+e^4 \log (x)+\left (6+6 e^4+\left (1+7 e^4\right ) \log (x)+e^4 \log ^2(x)\right ) \log \left (\frac {5+\log (2)}{6+\log (x)}\right )}{\left (6 x+\left (x+6 e^4 x\right ) \log (x)+e^4 x \log ^2(x)\right ) \log \left (\frac {5+\log (2)}{6+\log (x)}\right )} \, dx=\log \left (e^{4} \log \left (x\right ) + 1\right ) + \log \left (x\right ) - \log \left (\log \left (\frac {\log \left (2\right ) + 5}{\log \left (x\right ) + 6}\right )\right ) \]
integrate(((exp(4)*log(x)^2+(7*exp(4)+1)*log(x)+6*exp(4)+6)*log((log(2)+5) /(log(x)+6))+exp(4)*log(x)+1)/(x*exp(4)*log(x)^2+(6*x*exp(4)+x)*log(x)+6*x )/log((log(2)+5)/(log(x)+6)),x, algorithm=\
Time = 0.18 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \[ \int \frac {1+e^4 \log (x)+\left (6+6 e^4+\left (1+7 e^4\right ) \log (x)+e^4 \log ^2(x)\right ) \log \left (\frac {5+\log (2)}{6+\log (x)}\right )}{\left (6 x+\left (x+6 e^4 x\right ) \log (x)+e^4 x \log ^2(x)\right ) \log \left (\frac {5+\log (2)}{6+\log (x)}\right )} \, dx=\log {\left (x \right )} + \log {\left (\log {\left (x \right )} + e^{-4} \right )} - \log {\left (\log {\left (\frac {\log {\left (2 \right )} + 5}{\log {\left (x \right )} + 6} \right )} \right )} \]
integrate(((exp(4)*ln(x)**2+(7*exp(4)+1)*ln(x)+6*exp(4)+6)*ln((ln(2)+5)/(l n(x)+6))+exp(4)*ln(x)+1)/(x*exp(4)*ln(x)**2+(6*x*exp(4)+x)*ln(x)+6*x)/ln(( ln(2)+5)/(ln(x)+6)),x)
Time = 0.30 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.20 \[ \int \frac {1+e^4 \log (x)+\left (6+6 e^4+\left (1+7 e^4\right ) \log (x)+e^4 \log ^2(x)\right ) \log \left (\frac {5+\log (2)}{6+\log (x)}\right )}{\left (6 x+\left (x+6 e^4 x\right ) \log (x)+e^4 x \log ^2(x)\right ) \log \left (\frac {5+\log (2)}{6+\log (x)}\right )} \, dx=\log \left ({\left (e^{4} \log \left (x\right ) + 1\right )} e^{\left (-4\right )}\right ) + \log \left (x\right ) - \log \left (-\log \left (\log \left (2\right ) + 5\right ) + \log \left (\log \left (x\right ) + 6\right )\right ) \]
integrate(((exp(4)*log(x)^2+(7*exp(4)+1)*log(x)+6*exp(4)+6)*log((log(2)+5) /(log(x)+6))+exp(4)*log(x)+1)/(x*exp(4)*log(x)^2+(6*x*exp(4)+x)*log(x)+6*x )/log((log(2)+5)/(log(x)+6)),x, algorithm=\
Time = 0.34 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {1+e^4 \log (x)+\left (6+6 e^4+\left (1+7 e^4\right ) \log (x)+e^4 \log ^2(x)\right ) \log \left (\frac {5+\log (2)}{6+\log (x)}\right )}{\left (6 x+\left (x+6 e^4 x\right ) \log (x)+e^4 x \log ^2(x)\right ) \log \left (\frac {5+\log (2)}{6+\log (x)}\right )} \, dx=\log \left (e^{4} \log \left (x\right ) + 1\right ) + \log \left (x\right ) - \log \left (-\log \left (\log \left (2\right ) + 5\right ) + \log \left (\log \left (x\right ) + 6\right )\right ) \]
integrate(((exp(4)*log(x)^2+(7*exp(4)+1)*log(x)+6*exp(4)+6)*log((log(2)+5) /(log(x)+6))+exp(4)*log(x)+1)/(x*exp(4)*log(x)^2+(6*x*exp(4)+x)*log(x)+6*x )/log((log(2)+5)/(log(x)+6)),x, algorithm=\
Time = 13.30 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {1+e^4 \log (x)+\left (6+6 e^4+\left (1+7 e^4\right ) \log (x)+e^4 \log ^2(x)\right ) \log \left (\frac {5+\log (2)}{6+\log (x)}\right )}{\left (6 x+\left (x+6 e^4 x\right ) \log (x)+e^4 x \log ^2(x)\right ) \log \left (\frac {5+\log (2)}{6+\log (x)}\right )} \, dx=\ln \left ({\mathrm {e}}^{-4}+\ln \left (x\right )\right )-\ln \left (\ln \left (\frac {\ln \left (2\right )+5}{\ln \left (x\right )+6}\right )\right )+\ln \left (x\right ) \]